Before working Exercise, read the Guidelines in the Computational Exercises of Chapter 5,...

Before working Exercise, read the Guidelines in the Computational Exercises of Chapter 5, which discuss the Schrödinger equation adapted to numerical solution, with m and ii both defined as 1. Choosing only ψ(0) and ψ(Δx),the following relationship gives ψ(2Δx), then, by reapplication, ψ at every multiple of Δx.

ψ(2Δx) = 2ψ(Δx) - ψ(0) + 2Δx2(U(Δx) - E)ψ(Δx)

Here we consider only U(x) that are symmetric about x = 0, so that ψ(x) must be either an odd or even function of x. Our potential energy will model a one-dimensional crystal as an array of finite wells. A flexible (though not terribly elegant) way to define this U(x) starts by defining a "single well function" that is I everywhere except in a unit-width region centered on X, where it is B, standing for bottom.

sw(.x, X, B) - (B + l)/2 + sign[l/4 - (X - x)^2]*(B – 1)/2

If x differs from X by less than 1/2, this function is B, and if by more than 1/2, it is 1. Now we can define a U(x) that is U0 everywhere but in wells centered on x = 0, X1 X2, and X3, where it is 0.

U(x) = U0 sw(x, 0,0) sw(x, X,, 0) sw(x, X2,0) sw(x, X3, 0)

The assumption of symmetric U(x) allows us to plot \b for positive x only (we know it is odd or even), but it also means that, in effect, our crystal has three more wells at negative x, so it has seven atoms. (Note: Many further studies are possible within the framework provided. Exercise 80, for example, can investigate various atomic spacings, showing the effect on band width. The number of atoms can also easily be varied.)

Formation of Bands: Define U(x) with X1 X2, and X3 equal to 1, 2, and 3, respectively. This gives unit-width wells separated by walls of zero width—that is, one large well 7 units wide. For U0, use 20, and for Δx, use 0.001. (a) Following the Chapter 5 guidelines on choosing ψ(0) and ψ(Δx), test both odd and even functions at different trial values of E by finding \b at all multiples of Ax and plotting the results from x = 0 to x = 5. Find 14 allowed energies. Note that the indicator of having passed an allowed energy is the flip of the diverging large-x tail. The lowest energy or two will take the most work. Except for these, there is no need to exceed three significant figures, (b) With in and ћ both defined as 1 particle-in-a-box energies are simply n2π2/2L2. How do your energies compare, and why do some agree better than others? (c) Change X1 X2, and X3 to 1.2, 2.4, and 3.6, respectively, which gives seven equally spaced wells separated by walls of width 0.2. Again find 14 energies. Afterward, make a scatter plot of En versus n, where n goes from I to 14. Also make one of your part (a) results, then describe the main difference, (d) Consider the wave functions at the top of the first band and the bottom of the second. Kinetic energy depends on momentum, and thus on the function's wavelength (misshapen though it may be). Potential energy depends on the amplitude of the wave function in the walls, because these are the only places where our U(x) is nonzero. (If a wave function were always zero whenever U(x) is nonzero, the expectation value of U would be zero.) Compare the approximate kinetic-energies of these two states, then discuss quantitatively their potential energies. Argue that there should be a relatively large energy jump from one state to the other.